Problem: Layla has a coin that has a $60\%$ chance of showing heads each time it is flipped. She is going to flip the coin $5$ times. Let $X$ represent the number of heads she gets. What is the probability that she gets more than $3$ heads? You may round your answer to the nearest hundredth. $P(X>3)=$
Solution: Strategy (without a fancy calculator) The probability that Layla gets more than $3$ heads in the $5$ flips is equivalent to the probability that Layla gets $4$ or $5$ heads. So we can find those probabilities and add them them together to get our answer: $\begin{aligned} P(X>3)&=P(4\text{ heads})+P(5\text{ heads}) \\\\ &=P(X=4)+P(X=5) \end{aligned}$ Finding $P(X=5)$ For each flip, we know $P({\text{heads}})={60\%}$. To find the probability that all $5$ flips are heads, we can multiply probabilities since flips are independent: $\begin{aligned} P(X=5)&=({0.60})({0.60})({0.60})({0.60})({0.60}) \\\\ &=({0.60})^5 \\\\ &=0.07776 \end{aligned}$ We'll come back and use this result later. Next, we need to find $P(X=4)$ (the probability that she gets $4$ heads). Finding $P(X=4)$ Getting $4$ heads in $5$ attempts means Layla needs to get $4$ heads and $1$ tail. For each flip, we know $P({\text{heads}})={60\%}$ and $P({\text{tails}})={40\%}$. Let's start by finding the probability of getting $4$ heads followed by $1$ tail: $P({\text{HHHH}}{\text{T}})=({0.6})^4({0.4})=0.05184$ This isn't the entire probability though, because there are other ways to get $4$ heads from $5$ flips (for example, THHHH). How many different ways are there? We can use the combination formula: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _5\text{C}_4&=\dfrac{5!}{(5-4)!\cdot4!} \\\\ &=\dfrac{5 \cdot \cancel{4 \cdot 3 \cdot 2 \cdot 1}}{(1) \cdot \cancel{4 \cdot 3 \cdot 2 \cdot 1}} \\\\ &=5 \end{aligned}$ There are $5$ ways to get $4$ heads and $1$ tail. Do they all have the same probability? Each of the $5$ ways has the same probability that we already found: $\begin{aligned} P({\text{HHHH}}{\text{T}})&=({0.6})^4({0.4})=0.05184 \\\\ P({\text{HHH}}{\text{T}}{\text{H}})&=({0.6})^4({0.4})=0.05184 \\\\ P({\text{HH}}{\text{T}}{\text{HH}})&=({0.6})^4({0.4})=0.05184 \\\\ P({\text{H}}{\text{T}}{\text{HHH}})&=({0.6})^4({0.4})=0.05184 \\\\ P({\text{T}}{\text{HHHH}})&=({0.6})^4({0.4})=0.05184 \end{aligned}$ So we can multiply this probability by $5$ since that is how many ways there are to get $4$ heads in $5$ flips. $\begin{aligned} P(X=4)&=5(0.6)^4(0.4) \\\\ &=5(0.05184) \\\\ &=0.2592 \end{aligned}$ Putting it all together Let's return to our original strategy to answer the question: $\begin{aligned} P(X>3)&=P(4\text{ heads})+P(5\text{ heads}) \\\\ &=P(X=4)+P(X=5) \\\\ &=5(0.6)^4(0.4)+(0.6)^5 \\\\ &=0.2592+0.07776 \\\\ &=0.33696 \\\\ &\approx0.34 \end{aligned}$ The answer $P(X>3)=0.33696\approx0.34$